Find the sum of the first fifty terms of the sequence (an) by the given formula (an) = 4n-2.

First, we prove that the sequence (an, where n is a natural number) given by the formula an = 4 * n – 2 is an arithmetic progression.
Let’s apply the method of mathematical induction.
On the one hand: the first term is a1 = 4 * 1 – 2 = 2, the second term is a2 = 4 * 2 – 2 = 6, then the difference (step) is d = a2 – a1 = 6 – 2 = 4.
On the other side. Let n be any natural number. Then, if the number an is the nth term of an arithmetic progression, then an + 1 is the (n + 1) th term of the same arithmetic progression with the same step d = 4. Indeed, an + 1 – an = (4 * ( n + 1) – 2) – (4 * n – 2) = 4 * n + 4 – 2 – 4 * n + 2 = 4 = d.
Now we can calculate the sum of the first fifty members of the sequence. For this purpose, we will use the formula for determining the sum of the first n terms (Sn) of the arithmetic progression Sn = ((2 * a1 + d * (n – 1)) / 2) * n.
For n = 50, we have S50 = ((2 * 2 + 4 * (50 – 1)) / 2) * 50 = 100 * 50 = 5000.
Answer: 5000.